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Aim: How can we solve projectile motion (trajectory) problems?

Aim: How can we solve projectile motion (trajectory) problems?. Do Now: An object is thrown straight up with a velocity of 10 m/s. How long is it in the air for?. v i = 10 m/s a = -9.8 m/s 2 d = 0 m t = ?. d y = v iy t + ½a y t 2 t = 2 s. Projectiles. Launched at an angle. M.

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Aim: How can we solve projectile motion (trajectory) problems?

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  1. Aim: How can we solve projectile motion (trajectory) problems? Do Now: An object is thrown straight up with a velocity of 10 m/s. How long is it in the air for? vi = 10 m/s a = -9.8 m/s2 d = 0 m t = ? dy = viyt + ½ayt2 t = 2 s

  2. Projectiles Launched at an angle

  3. M

  4. http://www.youtube.com/watch?v=VgqE87UmQ10

  5. Evil Knievel

  6. Water Slide

  7. Projectiles have both x and y components

  8. Horizontal motion: Velocity is constant (the same as horizontally fired objects) Vertical Motion The same as vertically fired objects Physlet Physics Ch. 3 I.3.4 Vertical motion – look at the dots Horizontal motion – look at the dots

  9. Initial Velocity • Unlike horizontal projectiles, there are both vix and viy • Initial velocity can be resolved into x and y components vi viy= visinθ θ vix = vicosθ

  10. Can’t remember your trig?! • It’s in your reference table!!

  11. Is there an easier way to remember if vix or viy is sinθ or cosθ? • It’s in your reference table!! In this case, A = vi

  12. Now what? • Since you can determine your vix and viy, the rest we can figure out!!! • Horizontal velocity is still constant • Horizontal acceleration is still zero • Vertical acceleration is -9.81 m/s2 (just like in vertically fired objects) • Vertical vf at max height is 0 m/s (just like in vertically fired objects) • Vertical displacement is 0 m (the object starts on the ground and ends on the ground) • Time to max height = ½ total time

  13. ExampleYou kick a soccer ball into the air at a 30o angle with a velocity of 30 m/s. Find the time in the air. x y a = -9.8 m/s2 d = 0 m t = ? t = ? First step: Find vix and viy

  14. viy = visinθ viy = (30 m/s)sin30 viy = 15 m/s vix = vicosθ vix = (30 m/s)cos30 vix = 26 m/s x y a = -9.8 m/s2 d = 0 m t = ? viy = 15 m/s t = ? vix = 26 m/s d = vit + ½at2 0 m = (15 m/s)t + ½(-9.8 m/s2)t2 0 = 15t – 4.9t2 4.9t2 = 15t 4.9t = 15 t = 3.1 s

  15. Find the maximum height x y Choose a formula: vfmax2 = viy2 + 2admax dmax = viytmax + ½atmax2 a = -9.8 m/s2 d = 0 m viy = 15 m/s t = 3.1 s dmax = ? vfmax = 0 m/s tmax = 1.55 s vix = 26 m/s t = 3.1 s dmax = viytmax + ½atmax2 dmax = (15 m/s)(1.55 s) + ½(-9.8 m/s2)(1.55 s)2 dmax = 11.48 m

  16. Solve for the range Remember: vix = constant Therefore, vix = x y a = -9.8 m/s2 d = 0 m viy = 15 m/s t = 3.1 s dmax = ? vfmax = 0 m/s tmax = 1.55 s vix = 26 m/s t = 3.1 s d = ? d = 80.6 m

  17. Thinking question • What angle gives the greatest range? 45o

  18. Which angle gives the: greatest range? greatest height? 45° 90°

  19. Another example problem • A projectile is fired at an angle of 22o with an initial velocity of 120 meters/second. • How highwill the projectile travel? • How far will the projectile go?

  20. Pages 130-131

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